A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontal speed v. [a]. find the normal force between the sphere and the particle just after the impulse. [B]. What should be the minimum value of v for which the particle does not slip on the sphere?[ c]. Assuming the velocity v to be half the minimum calculated in part, [d]. find the angle made by the radius through the particle with the vertical when it leaves the sphere.

Explanation

a. Radius=R, horizontal speed=v
From the free body diagram
`mg-N=(mv^2)/(R)`
`impliesN=mg-(mv^2)/(R)`

When the particle is given maximum velocity so that the particle does not slip on the sphere.
`Nge0`
`(mv^2)/(R)=mg`
`v=sqrt(gR)`

c. If the body is given velocity `v_1` at the top such that
`v_1=(sqrt(gR))/(2)impliesv_1^2=(gR)/(4)`
Let the velocity be `v_2` when it leaves contact with the surface
So, `(mv_2^2)/(R)=mg cos theta`
`implies v_2^2=Rgcostheta` (i)
Again `(1/2)mv_2^2-(1/2)mv_1^2=mgR(1-costheta)`
`impliesv_2^2=v_1^2+2gR(1-costheta)` (ii)
From equation (i) and (ii),
`Rgcostheta=((Rg)/(4))=2gR(1-costheta)`
`implies cos theta=(1/4+2-2cos theta)`

`implies3costheta(9/4)`
`implies theta=cos^-1(3/4)`